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Albedo is the fraction of incident electromagnetic radiation power reflected, summed over the outgoing directions included in the measurement. For incident power $P_{\mathrm{in}}>0$ and reflected power $P_{\mathrm{ref}}$,Thus $\alpha =0.30$ means 30 percent reflected. For an opaque surface the remaining 70 percent is absorbed; for a transmitting layer, transmission must also be accounted for. Specify the incident wavelengths and illumination geometry: a visible-band surface albedo and a sunlight-weighted broadband surface albedo need not agree. NASA's definition explicitly allows a specified spectral band. ^[ref]

For a whole planet, broadband Bond albedo sums reflected power over all outgoing directions and all incident wavelengths, weighted by the illuminating spectrum. It is the fraction needed for a whole-planet absorbed-sunlight budget. A local surface albedo describes one surface under specified illumination; it cannot simply replace the Bond albedo of Earth, whose clouds and terrestrial atmosphere also reflect and absorb sunlight. Nor is Bond albedo the same as geometric albedo, which compares brightness in a particular viewing geometry. NASA's account of Earth's budget separates reflection by clouds, particles and the surface from absorption within the terrestrial atmosphere and at the surface. ^[ref]

The usefulness of albedo is that it turns optical behavior into an energy budget. Bright snow and sea ice usually absorb less sunlight than exposed water under comparable illumination, but their actual surface albedos depend on the spectrum and surface conditions. The ice-ocean albedo feedback makes that optical property part of the dynamics, while the following zero-dimensional planetary energy balance isolates the geometry and competing thermal radiation.

Altitude $\gamma$, also called elevation angle, is the signed angle from the astronomical horizon to a viewing direction, measured in the vertical plane containing that direction and the zenith. Its range is $-90^{\circ }\le \gamma \le 90^{\circ }$: positive altitude is above the horizontal plane, zero altitude lies on the astronomical horizon, and $90^{\circ }$ points to the zenith. Altitude measures direction, rather than a Euclidean distance or a height in metres. In the local horizontal coordinate system, $\sin \gamma =\mathbf{n}\cdot {\mathbf{e}}_U$; the dot product makes the sign unambiguous even for directions below the astronomical horizon.

A concrete motivating observation is the Exploratorium Teacher Institute's Height Sight activity. Its inclinometer uses a sighting tube fixed to a card and an angular scale, with a string and small weight hanging from the scale's pivot. A partner reads where the freely hanging string meets the scale. Sighting a level target should give a readout near zero; sighting a taller target gives an angular height. The activity asks the observer to move closer and farther: the closer position gives the larger angle. The interpretation is that the readout measures the tube's tilt relative to the gravity reference, while converting that angle to a physical height also requires distance and eye height. The activity also extends the inclinometer to sky observations. ^[ref].

The following calibration and sky sequence is a proposed extension, with no claimed observing dataset. First establish two reference marks at the same height with a water level, rather than assuming that a sloping floor or distant ridge is horizontal. Keep the inclinometer in a vertical plane, sight between the marks, wait for the weight to settle, and read the zero several times. Next sight a mark whose independently measured rise $q$ above the tube and horizontal separation $d$ are known; a right triangle predicts ${\gamma }_{\mathrm{ref}}=\arctan (q/d)$. A synthetic $q=3\mathrm{m}$, $d=5\mathrm{m}$ gives ${\gamma }_{\mathrm{ref}}=30.964^{\circ }$. Checking both zero and this nonzero angle can reveal a wrongly attached tube or a scale read from the wrong end. Use level separation, not distance along inclined ground.

If the scale's indicated altitude is $m={\gamma }_{\mathrm{app}}+b$, estimate its additive zero $b$ from the level sight, then obtain apparent altitude ${\gamma }_{\mathrm{app}}=m-b$. For synthetic readings $b=1.0^{\circ }$ and $m=32.0^{\circ }$, the corrected altitude is $31.0^{\circ }$, close to the $30.964^{\circ }$ calibration prediction. Declaring illustrative readout bounds of $0.5^{\circ }$ for each reading gives a conservative combined bound of $1.0^{\circ }$ on $m-b$; these bounds are assumptions, not measured performance. Repeatability tests the variation between readings, whereas a persistent zero offset remains a systematic measurement error. Include uncertainty in the reference rise and separation before judging agreement.

At a fixed site, sight a recognizable star, record the observation time, corrected altitude and north-through-east azimuth, and repeat over a short interval. Record the terrain boundary at the same azimuth separately from the gravity-defined zero. The fixed-direction model in horizontal coordinate system predicts changing angular readouts as the Earth rotates; this is a prospective test, rather than a fabricated measurement curve. Keep the tube aligned, the string free of contact and shielded from wind, the eye at the same reference position, and repeat the level check afterward. Timestamp each sight so genuine sky motion is not mistaken for poor repeatability. An optical telescope with an angular readout can replace the sighting tube, but its mounting still needs an independently established level and angular zero.

The corrected instrument reads apparent altitude, including atmospheric refraction. Under ordinary conditions atmospheric refraction raises a low direction, so if $R$ is the upward angular correction, ${\gamma }_{\mathrm{app}}={\gamma }_{\mathrm{geom}}+R$. Subtracting the instrument zero does not subtract $R$. Record site height, temperature and pressure if applying a refraction model: these are explicit observer inputs in the US Naval Observatory's NOVAS software. A coarse inclinometer may be unable to resolve the correction, and a local weather reading alone does not determine every atmospheric layer. ^[ref].

At a sightline perpendicular to Earth's instantaneous orbital velocity, a heliocentrically transverse ray acquires an apparent angular displacement towards the observer's motion. A telescope of length $L$ moves transversely by $v_{\oplus }L/c$ during the light crossing, giving the first-order result. The exact transverse Lorentz boost gives $\sin \delta =v_{\oplus }/c$. For a circular orbital radius $R$ and year $Y$, $v_{\oplus }=2\pi R/Y$. The amplitude is half the full opposite-season excursion for a suitable sightline; it is not annual parallax, which depends on stellar distance.

Aperture synthesis reconstructs the sky's brightness from complex visibilities measured at different interferometric baselines. Each measurement constrains a Fourier transform component of the brightness, rather than supplying a little photograph. Combining many such components can distinguish sources that each individual antenna leaves unresolved. The remarkable economy is that changing separations and the rotation of Earth can provide many measurements with only a few physical antennas.

Martin Ryle's Cambridge group made aperture synthesis a practical method of astronomical imaging at the Mullard Radio Astronomy Observatory. Cambridge's institutional history connects that work with the One-Mile Telescope and its larger successors.^[ref]

In his 1974 lecture Radio Telescopes of Large Resolving Power, Martin Ryle recalled the problem of distinguishing weak extragalactic sources and resolving their structure. He explained sequential wavefront sampling using one simultaneously observed element as a wave phase reference, and the saving obtained by letting Earth rotate the apparent baseline.^[ref]

The 1974 Nobel Prize in Physics distinguished Martin Ryle's observations and inventions, especially aperture synthesis, from Antony Hewish's role in discovering pulsars. The latter is a separate astronomical discovery, with its own observational history.^[ref]

The independent 5–50-day survey uses injected transits to measure completeness and reports a flattened occurrence distribution below about two Earth radii. Its period selection differs from the exam's 85-day cut; it is not treated as the source of the printed histogram.

A temporary reversal in a planet's apparent direction across the background sky, arising from the observer's changing position. For an outer planet observed from the Earth, the Earth overtakes it along its smaller, faster orbit. Different apparent reversals for two planets can produce three apparent conjunctions around one Sun-centred encounter, even when both follow prograde circular orbits.

Aristarchus of Samos used the half-lit Moon to estimate the relative distances of the Sun and Moon. In the spherical illumination model a half-lit disk means phase angle $\alpha =\pi /2$, so the Sun–Moon–Earth triangle is right-angled at the Moon. Keeping the Sun at a finite distance, let $d_{EM}$ and $d_{ES}$ be the distances from Earth to the Moon and Sun. Then elongation $ϵ$ at Earth givesThe $87^{\circ }$ elongation attributed to Aristarchus of Samos gives a ratio about $19.1$, consistent with his conclusion that the Sun was between $18$ and $20$ times farther away. The example shows how lunar phase geometry becomes a distance measurement, but the ratio is extremely sensitive to an angle close to $90^{\circ }$. Unlike the distant-Sun approximation, this finite triangle has elongation slightly less than $90^{\circ }$ at half illumination. ^[ref].

Astrometry measures the positions and motions of celestial objects. Repeated position measurements can distinguish stellar parallax, caused by the observer's motion, from proper motion of the source across the sky.

An astrometry measurement needs a date and a reference frame, as well as an angle. In an equatorial coordinate system, right ascension and declination label two viewing directions. A catalogue's reference date states when its quoted position applies; comparing entries at different dates without accounting for proper motion can make the same star appear misplaced. The observational problem is to infer a consistent history from successive incoming light directions, while accounting for the observer's changing position and measurement error.

To see the separation clearly, consider just one angular coordinate $\theta$, with $t$ the numerical elapsed time in years. Assume a simplified circular orbit of the Earth, a fixed source Euclidean distance, constant proper motion in this coordinate and no other accelerations. Choose the direction and the starting date so that the annual stellar parallax projection is the cosine termHere ${\theta }_0$ is the coordinate without the annual stellar parallax offset at the reference date; the observed $\theta (0)$ is ${\theta }_0+p$. The proper motion coefficient $\mu$ is quoted in arcseconds per year, and $p$ is the annual stellar parallax in arcseconds. This is an illustrative choice of a full-amplitude projection. General sky directions have different projected amplitudes and phases in their two coordinates; their apparent paths need not be circles. Real astrometry uses the observer's calculated position rather than imposing this single cosine. ^[ref].

Set ${\theta }_0=0$, $\mu =100\mathrm{mas}\mathrm{y}{\mathrm{r}}^{-1}$ and $p=50\mathrm{mas}$, where $1\mathrm{mas}$ is one thousandth of an arcsecond. The model gives $\theta (0)=50$, $\theta (0.5)=0$ and $\theta (1)=150\mathrm{mas}$. Same-season observations one year apart differ by $100\mathrm{mas}$ because the annual stellar parallax contribution repeats. Subtracting that fitted straight trend leaves an annual signal between $-50$ and $+50\mathrm{mas}$. The existing stellar parallax distance relation then gives $20$ parsecs.

Observing only at the same annual phase would recover the proper motion trend but leave $p$ indistinguishable from a constant offset. Sampling several seasons as well as several years separates the repeating signal from the accumulating displacement. The two signatures answer different questions using the same measurements.

The astronomical horizon is the great circle where the horizontal plane through the observer intersects the celestial sphere. It is perpendicular to the local vertical, so every direction on the astronomical horizon has altitude $0^{\circ }$ and zenith distance $90^{\circ }$. In the local horizontal coordinate system, it consists of the unit vectors satisfying $\mathbf{n}\cdot {\mathbf{e}}_U=0$. This geometric reference remains defined when buildings or mountains hide it; a visible skyline is not a calibration of zero altitude.

Three boundaries must be distinguished. The astronomical horizon fixes the angular reference. Terrain supplies a site-dependent boundary with apparent altitude $t(\eta )$ that changes with azimuth. Atmospheric refraction changes the apparent direction of incoming light, and can also change the apparent distant boundary. A point source clears an opaque terrain boundary only if ${\gamma }_{\mathrm{app}}>t(\eta )$ when both are expressed in the same apparent horizontal coordinate system; this clearance condition does not guarantee visibility through clouds or bright sky. In the synthetic diagram, a source with geometric altitude $3^{\circ }$ and an imposed illustrative upward correction $R=0.2^{\circ }$ remains behind a $4^{\circ }$ boundary. The chosen $R$ is an algebraic illustration, not a fitted atmospheric prediction.

Even an unobstructed sea boundary need not coincide with the astronomical horizon. On a spherical Earth of radius $r$, an observer at height $q$ sees the unrefracted tangent to the surface at a downward angle $D$ satisfying $\cos D=r/(r+q)$. The right-angle tangent triangle supplies this relation; the visible boundary then has altitude $-D$. With illustrative $r=6371\mathrm{km}$ and $q=10\mathrm{m}$, $D\simeq 0.1015^{\circ }$, before atmospheric refraction. Raising the eye therefore changes the visible boundary without redefining the horizontal plane.

The US Naval Observatory's rise/set definitions explicitly distinguish the geometric reference from apparent contact with a visible disk: the standard Sun calculation puts its centre $50^{\prime }$ below the horizontal plane, combining an average $16^{\prime }$ radius with $34^{\prime }$ of atmospheric refraction. An angular minute is $1^{\prime }=1/60^{\circ }$ in the minute and second of arc convention. This is an operational average under stated conditions, rather than a universal zero of altitude. The same source identifies weather, local topography and observer height as limits on predicted rise/set time. An observation log should therefore retain the level reference, site and eye height, terrain profile, observation time, and whether each reported altitude is geometric or apparent. ^[ref].

The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

The Cavendish experiment makes the attraction predicted by Newton's law of universal gravitation measurable between laboratory masses. Two small test masses hang at the ends of a horizontal rod in a Cavendish torsion balance; nearby larger source masses exert a tiny torque. A thin suspension fiber supplies an opposing torque as it twists. Measuring its twist and calibrating its torsion constant lets an experimenter infer the gravitational force and, in a modern analysis, the gravitational constant. ^[ref]

The beautiful feature is that the two horizontal forces can cancel as vectors while their torques add. The instrument converts weak attraction into a rotation, with sensitivity set by the fiber's torsion constant. Earth's much larger downward gravitational weight is supported by the suspension rather than supplying the intended torque about its vertical axis. A laboratory measurement can therefore probe the same Newtonian gravitational field that governs the two-body problem.

Henry Cavendish's original paper was Experiments to determine the Density of the Earth, published in Philosophical Transactions, volume 88 (1798), pp. 469–526, and read on 21 June 1798. On p. 469 he credits John Michell with the apparatus idea and explains that Michell died before using it; the instrument reached Cavendish through Francis John Hyde Wollaston. Cavendish's stated aim was Earth's mean mass density. Calling the experiment a determination of the gravitational constant is a modern interpretation, not Cavendish's own formulation of his result. ^[ref]

The connection between Earth's mean mass density and the gravitational constant follows in a spherical, nonrotating model of Earth. With Earth mass $M_E$, radius $R_E$, and surface gravitational acceleration $g$, the spherical shell theorem and Newton's law of universal gravitation giveThus a local comparison of gravitational attractions, together with knowledge of Earth's size and surface gravitational acceleration, reaches a property of the entire planet. These are modern equations explaining the connection, not equations quoted from Cavendish's paper.

For a worked example, use an explicitly simplified two-near-pair model. Each test point mass is $m$, each source point mass is $M$, and each lever arm is $L$, so the rod's full length is $2L$. Neglect the rod's mass. At zero twist put the test masses at $(-L,0)$ and $(L,0)$, and the sources at $(-L,-r)$ and $(L,r)$. Here $r$ is the near-pair center separation. Ignore each source's attraction on the test mass at the opposite end. Evaluate the two near-pair forces at zero twist and treat their torque as constant for a small twist, requiring $L|\theta |\ll r$. This is a teaching model, not a precision model of Cavendish's actual apparatus.

By Newton's law of universal gravitation, each retained force has magnitude $F=GmM/r^2$. The left force points down and the right force points up. Using torque about the suspension axis, both contributions are counterclockwise when viewed from above:The net retained horizontal force is zero, yet the net torque is positive. Moving the sources to $(-L,r)$ and $(L,-r)$ reverses both forces and hence the torque. In a real apparatus, the opposite source also attracts each test mass; that correction is explicitly included in the university derivation. ^[ref]

For illustrative parameters, take $m=0.010\mathrm{kg}$, $M=1.0\mathrm{kg}$, $r=0.050\mathrm{m}$, $L=0.10\mathrm{m}$, and torsion constant $\kappa =1.0\times 10^{-8}\mathrm{N}m/rad$. Use $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, the central value in NIST's 2022 CODATA table; that table gives standard uncertainty $0.00015\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$. This is an input for the example, not a measurement made with these assumed parameters. ^[ref]This sub-nanonewton force produces a measurable twist in the Cavendish torsion balance. The displayed digits make the toy calculation reproducible; they do not describe experimental precision.

In 2014, Gabriele Rosi and colleagues used atom interferometry: two rubidium-87 clouds prepared by laser cooling in vacuum, $328\mathrm{mm}$ apart, received shared three-pulse Raman transitions spaced by $160\mathrm{ms}$. Fluorescence readout and paired-fringe ellipse fitting yielded differential wave phase. Twenty-four tungsten alloy cylinders, about $516\mathrm{kg}$ total, moved between C and F. Cloud subtraction rejected shared vibration; configuration subtraction isolated the source change.^[ref]

Calibration used timed flight and light-sheet crossings; mirror tilt compensated Earth rotation. Wave vector reversal rejected unwanted shifts. Modeling included source Newtonian gravitational potential, atom trajectories, cloud distributions and moving supports. Source positioning, mass density variation and cloud geometry limited the reported combined relative standard uncertainty to $150$ parts per million.^[ref]

To understand the subtraction without reproducing that full analysis, consider an ideal atom interferometer in uniform vertical acceleration during each pulse sequence. Let $k_{\mathrm{eff}}$ be its signed effective vertical wave vector component and $T$ the pulse spacing. Its acceleration-dependent wave phase is $\varphi \simeq k_{\mathrm{eff}}g{T}^2$; labels U and L denote upper and lower clouds. The double difference is thenA background field identical in C and F cancels even if it differs between the two clouds. In a fixed-geometry, linear-response model, write $S=G\mathcal{K}$, where $\mathcal{K}$ is calculated from the source mass density, calibrated pulse sequence and atom trajectories. Thus $G_{\mathrm{inferred}}=S/\mathcal{K}$. Measuring wave phase accurately is one part of the task; calculating $\mathcal{K}$ accurately is another. The real nonuniform field and finite clouds require the trajectory-dependent analysis described above rather than this uniform-field approximation.

Take a cold, unmagnetized, collisionless plasma with uniform equilibrium electron density $n_e$, fixed neutralizing ions, and a small electromagnetic wave that is a transverse wave. Cold means that thermal-motion corrections are neglected. For electron mass $m_e$, elementary charge magnitude $e$ and electric field $\mathbf{E}=\mathrm{Re}(\hat{\mathbf{E}}{e}^{-i\omega t})$, the linear Newton's second law equation is $m_e\dot{\mathbf{v}}=-e\mathbf{E}$. Thus $\hat{\mathbf{v}}=-ie\hat{\mathbf{E}}/(m_e\omega )$ and the current density is $\hat{\mathbf{J}}=i{n}_e{e}^2\hat{\mathbf{E}}/(m_e\omega )$. Writing the induced electric polarization as $\hat{\mathbf{J}}=-i\omega \hat{\mathbf{P}}$ gives a negative electric susceptibility. With vacuum permittivity ${ϵ}_0$,The plasma frequency ${\omega }_p$ controls the response. For $\omega >{\omega }_p$, the propagating branch has real refractive index $0<n<1$. For $\omega <{\omega }_p$, this homogeneous model has an evanescent wave, not an ordinary propagating wave with a negative real refractive index. In a slowly varying vertical profile, a normally incident ray approaches a turning region where ${\omega }_p(z)=\omega$; the geometrical-ray approximation fails right at that region. At oblique incidence the conserved horizontal wave vector changes the turning condition. A smooth density gradient can therefore return a skywave without a mirror or a layer of metal.

For a homogeneous propagating section, direct differentiation of the dispersion relation givesA phase velocity greater than $c$ does not send information faster than $c$. Here a narrow wave packet has group velocity $v_g$, while the full causal response must still respect special relativity. Even this stripped-down plasma is strongly dispersive near its plasma frequency. In the actual ionosphere, collisions produce absorption, and the magnetic field of Earth changes the response with direction and wave polarization; a single real scalar $n$ cannot capture all of that behavior.

The interface between Earth's metallic core and silicate mantle, approximately 2890 km beneath the surface. It is a major material interface and a thermal boundary across which core heat enters the mantle. Its temperature is distinct from mantle potential temperature because pressure and the lower thermal boundary layer matter.

Use local east, north, and up unit vectors $({\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U)$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. In a spherical-Earth approximation at geographical latitude $\phi$, the angular velocity has components $\mathbf{\Omega }=(0,\Omega \cos \phi ,\Omega \sin \phi )$. Expanding the cross product for relative velocity $(v_E,v_N,v_U)$ givesThese are Coriolis acceleration components, not the total acceleration. For horizontal motion, the horizontal part is $(f{v}_N,-f{v}_E)$, where the Coriolis parameter is $f=2\Omega \sin \phi$. Thus northward motion in the northern hemisphere has an eastward Coriolis acceleration, to the right of its velocity; in the southern hemisphere the horizontal turning is to the left. At the equator $f=0$, but the full Coriolis acceleration need not vanish: vertical or eastward motion still couples through $\Omega \cos \phi$.

For a numerical example, take the Earth rotation rate $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, geographical latitude $\phi =45^{\circ }$, and an initially northward relative velocity ${\mathbf{v}}_{\mathrm{rot}}=20{\mathbf{e}}_N\mathrm{m}{\mathrm{s}}^{-1}$. Then

For a local, short-time estimate with no other horizontal force and zero initial eastward velocity, hold the initial northward velocity in the leading term: $\Delta x_E\simeq \frac{1}{2}fu{t}^2$. At $t=10\mathrm{s}$ this gives about $0.1031\mathrm{m}$ eastward. It assumes nearly fixed local axes and geographical latitude, negligible changes of height, and $|ft|\ll 1$. It is not a complete wind trajectory: pressure gradients, friction, and evolving velocity must enter an atmospheric equation of motion. The effective gravity in a rotating frame already combines ordinary gravity with the steady terrestrial centrifugal acceleration.

At first order in Earth's rotation, vertical velocity gives east acceleration $-2\Omega \cos \lambda \dot{z}$. A launch returning to its initial height lands westward, while a drop from the launch's maximum height lands eastward with one quarter the displacement magnitude. These results assume constant effective gravity, no air drag and no horizontal velocity in the Coriolis term. The northern component vanishes at this order.

The Coriolis force is the apparent force proportional to a particle's velocity relative to a rotating reference frame. For constant angular velocity $\mathbf{\Omega }$ and particle mass $m$,Here $\mathbf{v}$ in the title formula means ${\mathbf{v}}_{\mathrm{rot}}$, not the velocity measured in an inertial frame. The Coriolis force belongs to the chosen reference frame: a particle stationary in that rotating reference frame has no Coriolis force, even though it can have a nonzero velocity in the laboratory. Other reference frames assign different angular velocities and relative velocities, and hence different Coriolis forces to the same motion.

The cross product makes the Coriolis force perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}_{\mathrm{rot}}$, with magnitude $2m\Omega |{\mathbf{v}}_{\mathrm{rot}}|\sin \alpha$, where $\alpha$ is the angle between them. Apply the right-hand rule to $\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$ and then reverse its direction. Motion parallel to the rotation axis gives zero Coriolis force; reversing the relative velocity reverses the Coriolis force.

The appeal of this description is practical: axes attached to the Earth or a turning apparatus can be kept fixed on the page, provided Newton's second law includes the extra terms. A free particle in rotating coordinates shows exactly how a straight laboratory trajectory can become curved, while Coriolis acceleration in local Earth coordinates gives a useful terrestrial scale. The rotating-table demonstration of Coriolis deflection makes the two viewpoints visible in one experiment.

Declination $\delta$ is the signed angle from the reference celestial equator to a direction, measured along its hour circle: $-90^{\circ }\le \delta \le +90^{\circ }$. Positive declination means celestial north, rather than necessarily north of the observer's zenith. A direction with $\delta =0$ lies on the celestial equator; $\delta =+90^{\circ }$ is the north reference celestial pole. Declination is an angle, not a height or a measure of Euclidean distance. ^[ref].

The connection to an observation follows from a dot product. In the elementary spherical-Earth, unrefracted model, an observer at geographical latitude $\varphi$ has zenith direction $\mathbf{z}=(\cos \varphi \cos \Theta ,\cos \varphi \sin \Theta ,\sin \varphi )$ in the same reference axes. Since $\mathbf{n}\cdot \mathbf{z}=\sin a$ for altitude $a$, the cosine addition formula givesAt upper meridian transit, $H=0$, so $a_{\mathrm{upper}}=90^{\circ }-|\varphi -\delta |$. For the synthetic site $\varphi =52^{\circ }$ and synthetic direction $\delta =20^{\circ }$, this yields $a_{\mathrm{upper}}=58^{\circ }$. A direction with $\delta =70^{\circ }$ instead reaches $72^{\circ }$, north of the zenith; positive declination does not imply a crossing on the southern half of the celestial meridian. A negative computed upper altitude means the direction never rises above the geometric astronomical horizon at that site.

For an angular readout, define signed zenith distance $z_s$ on the upper celestial meridian, positive toward the south and negative toward the north. Then $\delta =\varphi -z_s$. A synthetic southward readout $z_s=32^{\circ }$ at $\varphi =52^{\circ }$ recovers $\delta =20^{\circ }$. An unsigned distance $|z_s|$ alone would leave two possibilities, $\varphi \pm |z_s|$, when both are in range. A calibrated optical telescope must therefore record which side of the zenith it points toward. In this linear relation, a $1^{\prime }$ error in the assumed geographical latitude produces a $1^{\prime }$ error in inferred declination; averaging repeated readings cannot remove that shared offset. Near the astronomical horizon, uncorrected atmospheric refraction also invalidates the geometric readout.

The unit vector construction explains a subtle feature of sky maps. At fixed declination, changing right ascension traces a parallel whose radius on the unit sphere is $\cos \delta$. Its arc length is $\cos \delta |d\alpha |$, with angles in radians; at $\delta =60^{\circ }$, a $15^{\circ }$ right ascension increment spans a $7.5^{\circ }$ arc along that parallel. This arc is not generally the shortest great circle separation. At either celestial pole, all right ascension values represent the same direction: the coordinate singularity belongs to the coordinate grid. The unit vector remains regular there.

Earth-rotation aperture synthesis uses Earth's changing orientation to vary a fixed ground baseline's projection onto the plane of the sky. In two dimensions, $u$ and $v$ are the components of ${\mathbf{b}}_{\perp }/\lambda$ along two perpendicular sky tangents. Tracking one field traces a curve in this $(u,v)$ plane; changing the physical baseline supplies other curves. A track samples Fourier transform coordinates, rather than the brightness along a track across the sky.

For a horizontal east–west baseline of length $B$, choose the sky-axis signs so thatwhere $H$ is hour angle and $\delta$ is declination. The ellipse follows by projecting the rotating baseline onto sky tangents. At a celestial pole it is a circle; near the celestial equator its minor axis collapses. Thus the same ground layout constrains different sky directions unequally. Rotation improves coverage, but does not fill every spatial frequency, and synthesis across many hours assumes that the source changes negligibly during that interval.

Earth's axial rotation relative to an inertial frame. The Foucault pendulum detects its component along the local vertical through the Latitude law of Foucault precession.